Table of Contents

## How to correctly calculate the capacity of the gas boiler?

When designing a heating **boiler** manufacturers pay great attention to the power indicator. It is this parameter will further characterize the performance and productivity of the heater in the circuit. It is important for the consumer to understand how to calculate the capacity of the gas boiler and achieve optimal interaction of the heater with other elements of the circuit, because the key to the effective operation of the heating system is the compatibility and coherence of all components.

The basic techniques of calculating the capacity of the device will be covered in this review. In the course of acquaintance with them, the property owner will be able to navigate in the parameters of choosing heating equipment for his home, as well as learn to compare different boilers and their performance characteristics.

### Mistake: When selecting focus only on the area of the room

Very often sellers of heating equipment are primarily interested in the **area** of the room. Previously, with typical construction and uniform GOSTs, the size of the area was sufficient to assess the situation. After all, all houses were built on the same principle. from sand-lime brick, a thickness of not less than 51 cm, and the heat loss of such a room was 100 W per 1 m 2. So it turns out that the boiler standard capacity of 24 kW (24,000 W) is designed for a room area of 240 m2.

But this is today the **area** of the room. is not the only thing that should be of interest to a potential seller of equipment. Because nowadays not only standard sand-lime brick is used as a material, but also others: steam-therm, gas blocks, expanded clay blocks. And the frame-shielded houses are built completely out of insulation.

Power of modern boilers is prescribed by the manufacturer in kW, but not in the ability to heat a certain **area**.

Therefore it is necessary to calculate the heat load of the house, and in particular all the rooms separately for a more accurate selection of radiators. After all, any heating system is, above all, only heat devices that compensate for the leakage of positive temperature from the room to the atmosphere. And also provide circulation of fresh cold air through ventilation.

### Wrong : Leave the underfloor heating without considering it

Do not consider the capacity of the underfloor heating when choosing a boiler.A second serious mistake. Taking the area of the house, for example, 240m 2. The user gets the necessary output of 24 kW. But for some reason he also adds the floor heating capacity. Wrongly using only the floor **area**. Using a floor heating of 100 square meters. Adding an extra 10 kW to the floor heating. Getting a total of 34 kW.But the floor heating itself is source of heat. It covers part of the heating load, and the rest you just need to add radiators, because when we add a warm floor, heat loss in the house does not increase, they remain the same. 24 kW.

### Mistake : While calculating the necessary power to also take into account the reserve for the **boiler**.

Any modern **boiler**, whether it is a wall or floor boiler works not only as a space heater, but also for DHW. When there is a need for hot water from the faucet or heating water in the **boiler**, the boiler is fully switched to this function and at this time bears the load only for this purpose.

On average, the heat exchanger capacity of boilers is from 18 to 24 kW, depending on the model. But, nevertheless, many sellers mistakenly recommend the radiators capacity of 24 kW per room in 240 m 2, plus 10 kW on the floor heating, plus 20 kW on the boiler. The total capacity of the **boiler**, in their opinion, should be 54 kW. And this significantly increases the cost of the boiler, although you can do with less powerful, but less expensive equipment.

## Example of calculation of heat losses

As an example, let’s calculate the heat loss of a house with the given characteristics.

According to the design, the width of the building is 10 m, length 12 m, ceiling height 2.7 m, the walls are oriented north, south, east and west. In the west wall built in 3 windows, two of them have the dimensions of 1.5х1.7 м. one 0.6х0.3 м.

In the southern wall there are built-in doors with dimensions of 1.3×2 m, there is also a small window 0.5×0.3 м. On the east side there are two 2.1×1.5 m and one 1.5×1.7 м.

- Wall sheathing Fibreboard outside and inside 1.2 cm each, coefficient 0.05.
- glass wool placed between the walls, its thickness 10 cm and coefficient 0.043.

The thermal resistance of each wall is calculated separately, t.к. the structure in relation to the cardinal points of the compass, the number and the area of the openings are taken into account. The results of the wall calculations sum up.

The floor is multi-layered, over the entire area is made with the same technology, includes:

- Tongue and groove board with a thickness of 3.2 cm, coefficient of thermal conductivity 0.15.
- dry levelling layer 10 cm thick particleboard with a ratio of 0.15.
- Thermal insulation. mineral wool 5 cm thick, coefficient 0.039.

Let’s assume that the floor has no trapdoors to the basement or similar openings. Therefore the calculation is performed for the area of all rooms according to the same formula.

Let’s assume that the ceiling also has no access to the attic above the house or utility room.

The house is located in Bryansk region, in Bryansk city, where the critical negative temperature is.26 degrees. It was found experimentally that the ground temperature is 8 degrees. The desired room temperature is 22 degrees.

### Calculation of heat losses of walls

To find total thermal resistance of the wall you have to calculate thermal resistance of every layer of the wall.

A layer of glass wool is 10 cm thick. This value must be converted to meters, that is:

You have the value B=0.1. The thermal conductivity coefficient of the insulation is 0.043. Substitute the data in the formula for the thermal resistance and get:

Using a similar example, let’s calculate the resistance to heat of the isoply:

Total thermal resistance of the wall will be equal to the sum of thermal resistance of every layer, taking into account that we have two fiberboard layers.

Having determined total heat resistance of the wall, we can find heat losses. For every wall they are calculated separately. Let’s calculate Q for the north wall.

According to the plan, the north wall has no windows, its length is 10 m, height. 2.7 м. Then the area of the wall S is calculated by the formula:

Let’s calculate the parameter dT. It is known that the critical ambient temperature for Bryansk is 26 degrees, and the desired room temperature is 22 degrees. Then

For the north wall the additional factor L=1 is taken into account.1.

Having made preliminary calculations, you can use the formula to calculate heat loss:

Let’s calculate the heat loss for the western wall. Based on the data, there are 3 windows, two of them have dimensions of 1.5х1.7 m and one 0.6х0.3 м. Calculate the **area**.

From the total area of the west wall you must exclude the area of windows, because their heat loss will be different. To do this we calculate the area.

To calculate the heat losses we will use the **area** of the wall without window area, that is

For the west side the addition factor is 1.05. The data obtained are substituted into the basic formula for calculating heat losses.

Similar calculations do for the east side. There are 3 windows, one of them has the following dimensions.5х1.7 m, the other two are 2.1х1.5 м. Calculate their area.

From the total area of the wall subtract the window area values:

The addition factor for the east wall is.1.05. Based on the data, calculate the heat loss of the eastern wall.

On the southern wall there is a door with the parameters 1.3×2 m and a window 0.5х0.3 м. Let’s calculate their **area**.

Determine the area of the wall without taking into account the windows and doors.

Calculate the heat loss of the south wall with the coefficient L=1.

Having determined the heat losses of each of the walls, we can find their total heat losses by the formula:

As a result, the heat loss of the walls was 1,810 W per hour.

### Calculation of heat losses of windows

There are 7 windows in the house, three of them have size 1.5×1.7 m, two 2.1×1.5 m, one 0.6×0.3 m and another 0.5×0.3 м.

Windows with dimensions of 1.5×1.7 m is a double chamber PVC profile with I-glass. From the technical documentation we can learn that its R=0.53. Windows with dimensions 2.1×1.5 m double-glazed windows with argon and I-glass have thermal resistance R=0.75, windows 0.6х0.3 m and 0.5×0.3 R=0.53.

It is also important to take into account orientation of windows in relation to sides of the world.

Let’s calculate the heat losses of the western windows, taking into account the coefficient L=1.05. On the side there are 2 windows with overall dimensions 1.5×1.7 m and one with 0.6×0.3 м.

The total loss of the western windows is

On the south side there is a window 0.5×0.3, its R=0.53. Let’s calculate its heat loss with the coefficient of 1.

On the east side there are 2 windows with dimensions 2.1×1.5 and one window 1.5×1.7. Let’s calculate the heat losses taking into account the coefficient L=1.05.

Let’s sum up the heat losses of the east windows.

Total heat energy output through the windows is 1190 W.

### Determination of heat losses of doors

There is one door in the house, it is built into the south wall and has the dimensions of 1.3×2 м. Based on the passport data, the thermal conductivity of the door material is 0.14, its thickness 0.05 м. Due to these values it is possible to calculate the thermal resistance of the door.

To calculate we need to calculate its area.

After calculation of thermal resistance and area, we can find the heat losses. The door is placed on the south side, so we use an additional factor of 1.

Total, 347 W of heat comes out through the door.

### Calculation of the thermal resistance of the floor

According to the technical documentation, the floor is multilayered, the entire **area** is performed equally, has the dimensions of 10×12 m. Let’s calculate its area.

The floor consists of boards, particle board and insulation.

Thermal resistance has to be calculated for each floor layer separately.

Total thermal resistance of the floor is:

Considering that in winter the ground temperature is kept at 8 degrees, the temperature difference will be equal:

Using preliminary calculations you can find out house heat losses through the floor.

When calculating heat losses of the floor we take into account the coefficient L=1.

Total heat losses of the floor are 1885 W.

### Calculation of heat loss through the ceiling

When calculating the heat loss of the ceiling, a layer of mineral wool and wooden boards are taken into account. Vapour barriers and water proofing are not part of the thermal insulation, so we do not take them into consideration. For the calculations we need to find the thermal resistance of the wooden boards and the mineral wool layer. Let’s use their thermal conductivity coefficients and thickness.

The total thermal resistance will be equal to the sum of Rder.panel and Rmin.wool.

Then calculate the heat loss of the ceiling, taking into account the coefficient L=1.

In order to determine the total heat loss of the house we have to add up the heat losses of walls, windows, door, ceiling and floor.

To heat a house with the above parameters, you need a gas **boiler**, maintaining the power 8949 W or about 10 kW.

### Determination of heat loss including infiltration

Infiltration is a natural process of heat exchange between the external environment, which occurs during the movement of people in the house, when you open the front door, windows.

To calculate the heat loss for ventilation, you can use the formula:

- K is the design factor of air exchange ratio, for living rooms the coefficient 0.3, for heated rooms 0.8, for kitchen and bathroom 1.
- V volume of the room, calculated using height, length and width.
- dT the temperature difference between the ambient and the living space of the house.

### What size **boiler** do I need? (UK) | Boiler Size Calculator

A similar formula can be used if the room is ventilated.

Room height 2.7 m, width 10 m, length 12 m. Knowing this data, you can find its volume.

We take 0 as K coefficient.3. At that time

To the total calculated index Q it is necessary to add Qinf. As a result

Taking into account the infiltration the heat loss of the house will be 10489 W or 10.49 kW.

## What values are used in the calculations?

The simplest calculation of the boiler power according to the

areais as follows: take 1 kW power for every 10 sq.m. м. But it is worth remembering that these standards were drawn up in the Soviet Union. They do not take into account the modern construction technologies, moreover, they can be inadequate in an area where the climate is very different from Moscow and Moscow suburbs. Such calculations can be suitable for a small building with an insulated attic, low ceilings, excellent thermal insulation, windows with double glazing, etc.п. Alas, these requirements meet only a few buildings. To make a more detailed calculation of the boiler power, you need to consider a number of factors, such as:

- climatic conditions in the region;
- dimensions of the living space;
- The degree of insulation of the house;
- possible heat losses of the building;
- amount of heat required to heat water.

In addition, in homes with forced ventilation, the calculation of the **boiler** for heating must take into account the amount of energy needed to heat the air. As a rule, you need to use special software for calculations:

Carrying out the calculation of the capacity of the gas boiler, you should add about 20% more in case of emergencies, such as a cold spell or a drop in gas pressure in the system.

When designing the heating system is required to calculate not only the boiler capacity, but also the number of radiator sections. You can learn more about this in our article: https://aqua-rmnt.com/otoplenie/raschety/raschet-kolichestva-sektsiy-radiatorov-otopleniya.html.

## Calculation of boiler output by area

For a rough estimate of the required capacity of the heating unit is enough **area** of the premises. The simplest version for the middle zone of Russia assumes that 1 kW of power can heat 10m 2 square meters. If you have a house **area** of 160m2, the boiler capacity for its heating 16kW.

These calculations are approximate, since we don’t consider either the ceiling height or the climate. To this end, there are experimentally derived coefficients, by which the appropriate adjustments are made.

The given norm of 1 kW per 10 m2 is suitable for the ceilings of 2,5-2,7 m. If you have higher ceilings, you need to calculate the coefficients and recalculate. To do this, divide the height of your premises by the standard 2.7m and obtain the correction factor.

Calculating the output of the boiler by **area** is the easiest way

For example, the ceiling height of 3,2m. Let’s calculate the coefficient: 3.2m/2.7m=1.18 rounded up, we get 1.2. It turns out that for heating the premises 160m2 with ceiling height 3,2m requires a **boiler** capacity 16kW1,2 = 19,2kW. Usually they round up, so 20kW.

To consider the climatic features there are ready-made coefficients. For Russia they are as follows:

If the house is in the middle zone, just south of Moscow, apply a factor of 1.2 (20kW1.2 = 24kW), if in the south of Russia in the Krasnodar region, for example, a factor of 0.8, that is, the power required less (20kW0.8 = 16kW).

The calculation of heating and **boiler** selection is an important stage. Find the power incorrectly and you can get this result

These are the main factors to consider. But the found values are valid, if the boiler will work only for heating. If you also need to heat water, you need to add 20-25% of the calculated figures. Then you need to add a “margin” for peak winter temperatures. That’s another 10%. Total we get:

- For home heating and DHW in the middle belt 24kW20%=28.8kW. Then the reserve for the cold 28.8 kW10%=31.68 kW. Rounded and we get 32kW. If we compare with the original figure of 16kW, the difference is twice as much.
- The house in the Krasnodar region. Then we add the power for hot water heating: 16kW20%=19.2kW. Now the “reserve” for the cold 19,210% = 21,12 kW. Let’s round it up: 22kW. The difference is not so striking, but still quite decent.

From the examples it can be seen that even these values must be taken into account. But it is obvious that in calculating the boiler capacity for the house and apartment, there should be a difference. We can go the same way and use the coefficients for each factor. But there is a simpler way, which allows you to make adjustments at once.

When calculating the boiler heating for the house we apply a factor of 1.5. It takes into account the heat loss through the roof, the floor, the foundation. Fair enough for an average (normal) degree of insulation of walls masonry in two bricks or similar building materials.

For apartments other coefficients are used. If there is a heated room upstairs (another apartment) factor 0.7, if the heated attic 0.9, if unheated attic 1.0. You need to multiply the capacity of the boiler by one of these coefficients, calculated using the above method, and you will get a fairly reliable value.

To demonstrate the calculations, calculate the capacity of the gas heating boiler for an apartment 65 m2 with 3 m ceilings, which is located in the middle belt of Russia.

- Determine the required power on the basis of the area: 65 m2 /10 m2 =6,5 kW.
- Let’s correct for the region: 6,5 kW1,2=7,8 kW.
- The boiler will heat water, so add 25% (we like it hotter) 7,8 kW1,25=9,75 kW.
- Add 10% for the cold: 7,95kW1,1=10,725kW.

Now round up the result and you get: 11KW.

The above algorithm is true for the selection of heating boilers for any type of fuel. Calculating the capacity of an electric heating boiler will not differ from the calculation of solid fuel, gas or liquid fuel boiler. The main thing is the boiler capacity and efficiency, and heat loss from the boiler type does not change. The whole question is how to spend less energy. And this is the **area** of insulation.

## How to calculate the **boiler** capacity for heating?

Autonomous heating in a private home. it is convenient, affordable and generally effective. But only if the equipment is chosen correctly. We are talking not only about the types of boilers, but also the necessary capacity of the system. In this article we will share with you how to calculate the capacity of the boiler for heating, how to make the necessary calculations and whether it is worth doing it yourself.

- Boiler with too much power reserve. unjustified heating costs;
- Insufficient power. lack of heat in the apartment or private house.

Therefore, this issue should be approached with particular care.

A simplified scheme for determining energy costs is based on knowing the total **area** and the specific heat loss coefficient. In the middle zone, each square meter of the building loses 100 W, if the ceilings are up to 270 cm high. By multiplying the area of the building by the heat loss per cubic metre, 200 m²×100 W/m² = 20,000 W, we obtain the power needed to heat the house.

A similar calculation can be made knowing the volume of a residential building. This method is convenient when the height of the ceilings in the premises is different and differs from 270 cm. Specific heat loss per cubic metre is taken as 40 W/m³. Let’s assume a ceiling height of 325 cm throughout, then 200 m²×3.25 m×40 W/m³ = 26000 W would be needed to heat the house.

However, simplified calculations take little account of factors such as:

- climatic anomalies;
- additional energy costs associated with living in the house, such as hot water, shower, dryer, underfloor heating, garage, etc.д.;
- no heat map.

When building a new home, the last one can be made up. But when buying a home on the secondary market, there will be difficulties with this.

### Depending on the characteristics of the house

In the absence of project documentation is difficult to take into account its specifics in the calculations. Nevertheless, based on information from modern SNIP (building codes and regulations), as well as available to the owner, you can estimate the level of heat loss.

The parameters of the house that can be obtained:

- type of soil;
- Type of foundation and its arrangement;
- the value of the groundwater table;
- Frost depth,
- wall material;
- their thickness;
- presence of thermal insulation and its properties
- type of roof and ceiling.

### Power reserve for water heating

This factor significantly affects the choice of the number of kilowatts of **boiler**.

For a proper calculation it will be necessary to determine these parameters:

- Average daily consumption of hot water per resident;
- What technical means will be used for heating;
- Consumption dynamics, that is, the evenness of consumption during the day.

According to SNiP 2.04.01-85 in houses with showers and bathtubs (up to 170 cm in length), the average daily norm of hot water per resident is 115 liters. That is, for a family of 3 people, the DHW system should supply about 350 liters. If a two-circuit **boiler** with heating of flowing water will be used, then the capacity for heating should be increased by 30%. We obtain (20000 26000) W x 1,3 = (26000 33800) W for one bath.

It is customary to make 3 water points in the house (bath, washbasin, kitchen). And then the boiler power must be increased by 50%, which will give a new value of 30000-39000 W.

It is possible to reduce the requirements to the unit by installing a boiler or a gas column, but in the first case, it is not possible to obtain the necessary volume of water in a short time. An electric heater for 3000 W will need 40-50 minutes to raise the temperature of 100 liters of water to 55 ° C. In the second case, you will need to spend money on the purchase and installation of equipment, look for space in the house and the opportunity to organize a separate channel for smoke removal.

### In case of severe frost

The peculiarities of the regional climate also influence the choice of heating capacity. Most calculations are based on average values and do not take into account short-term fluctuations. However, if for 1-1,5 months or even for a week there will be frost of 10-20°C below the norm in this case it will be necessary to make allowance in calculations for such anomalies.

To the regional characteristics should be added a correction for wind strength and humidity. Some structures at.10°C and damp winds are colder than in.20°C and dry weather.

To compensate for extreme phenomena, up to 20% of design capacity is laid down, which is reflected in the value of climatic coefficients:

Continuing the estimate for a private home, we arrive at (30-38)×1.2=(36-46)kW.

## Floor standing gas boilers

The power produced by this type of equipment varies from ten to hundreds of kilowatts.

These models do not skimp on weight and use design solutions that guarantee reliability and durability. Lack of restrictions on weight and dimensions reduces their cost. Floor standing gas boilers use all new technological and design solutions. They can be combined and work together if there is a need for high power.

Such varieties of units are produced:

- Single-circuit. They are used only for heating.
- Two-circuit. They heat the space and provide DHW.
- Condensing. Can be 1-and 2-circuit, with an efficiency up to 109%.

When choosing the equipment in the floor version, the following characteristics must be taken into account

- energy dependence;
- material of heat exchangers;
- their number and location;
- characteristics of the gas removal system;
- type of combustion chamber;
- scheme of air supply;
- The way of ignition of the burner;
- degree of protection and automation;
- possibility of remote control and monitoring.

With floor standing boilers the problem of low pressure in the gas pipe is the easiest to solve.

A solid fuel combination boiler that can run on a variety of fuels. Reliable and inexpensive energy-independent steel boilers “Lemax Premium” offers the same name company from Taganrog.

### Buderus Logano G234, WS 38

Energy independent gas floor standing **boiler** from Germany with open burner at 38 kW. Electronic ignition, the heat exchanger made of cast iron. Gas consumption 4,17 m³/h (3,07 kg/h), from 215 W electric network. Height 120 cm, width 65 cm, depth 73 cm, weight 221 kg. Heating circuit pressure 4 bar. The design allows to connect the boiler. Warranty from the manufacturer for 2 years. Low nitrogen oxide level.

### Baxi slim EF 1.39

Italian energy-independent atmospheric gas boiler with a cast-iron heat exchanger of 39 kW. Gas consumption 4.73 m³/h (3.53 kg/h). Weight 150 kg, dimensions 85 x 40 x 67 cm. 2 year manufacturer’s warranty. Possibility to connect the boiler, optionally equipped with weather dependent automatics. Equipped with overheat protection, control of pressure in the heating circuit.

In the line of this manufacturer also has a gas boiler 30 kW Baxi Luna 3 310 FI.

### Navien GA-35KN

Two-circuit floor gas boiler of the Korean manufacturer, energy-independent, with a closed burner and DHW support, capacity 35 kW. Steel heat exchanger. Possibility to connect a boiler. Heating circuit range 2,9 0,2 bar. Gas consumption 3.34 m³/h (LPG 2.94 kg/h). Power Consumption. power 110W. Dimensions 85,6 x 40,2 x 63,1 cm, weight 86 kg. Equipped with LCD display and a water flow sensor. The manufacturer provides a 2 year warranty.

This manufacturer also has a gas boiler 40 kW Navien Deluxe-40K/

### ACV Alfa Comfort E 40

Energy-dependent floor standing gas boiler 32 kW made in. with an open injection burner and a cast iron heat exchanger. Working pressure of the heating circuit 0.8 to 3 bar. Maximum gas consumption 3.6 m³/h. Electric power consumption.130 W. Dimensions 97 x 51 x 57 cm, weight 136 kg. Maximum pressure in the heating circuit 4.5 bar at 90 °C. Has a control function for an external water heater with DHW priority. 5-year warranty.

### Protherm Bear 40 TLO

It is a floor convection boiler with a cast iron heat exchanger, open burner and natural smoke extraction of 35 kW. It is designed to work in conditions of complete absence of electricity and low pressure in the gas supply line. The **boiler** is ignited manually by means of a piezo element. Control unit operation is ensured by voltage from the thermocouple. Extinguishing of the flame leads to automatic gas shutoff.

Fuel consumption at maximum power is 4 m³/h. Dimensions 88 x 59 x 62 cm. Weight 130 kg.

## How to save energy costs?

We proceed from the fact that, firstly, the power consumption depends directly on the thermal capacity of the boiler. And secondly, most of the electricity consumed is taken by the circulating pump, which drives the coolant in the pipes, so that the pipes and radiators are heated dimensionally.

Let’s name a number of specific suggestions for those who would still like to reduce energy costs:

- Opt for an energy-independent unit. Most likely it will be a floor version. Alas, it is not able to compete with its power-dependent analogues in functionality and comfort.
- To buy the unit, which is independent, but of low power. Here, of course, there is a significant limitation. you can not ignore the number of heated square meters. If, for example, you need to heat 180-200 square meters of private homes, the gas boiler needs a capacity of 20-24 kW. And not less.
- Carefully study the product range of different brands. There are nuances inherent in each model and you may find the most attractive figures on electricity consumption in the technical specifications for some of them.
- Analyze what the total cost of electricity is made up of. Perhaps the share of these costs attributable to the gas boiler is negligible, and we need to shift attention to other facilities that are really over-consuming electricity.
- What about using alternative energy. let’s say solar panels or collectors on the roof of the house?

And yet in the pursuit of saving electricity, do not take your own actions to the point of absurdity. Do not forget that gas-fired units do not consume much electricity, because their main fuel resource is not electricity, but natural or liquefied gas.